Proof of $n - \lceil \dfrac{n}{2} \rceil = \lfloor \dfrac{n}{2} \rfloor$ (Latex test - MathJax)

The key here is to divide the cases, it is not really difficult at all

To proof : $n - \lceil \dfrac{n}{2} \rceil  = \lfloor \dfrac{n}{2} \rfloor$

Case 1.  $ n = 2k + 1$

$$n - \lceil \dfrac{n}{2} \rceil  = 2k + 1 - ( k+1) = \lfloor k + \dfrac{1}{2} \rfloor = k$$

Both the LHS and RHS are $k$ therefore, the equality stands true.

Case 2.  $ n = 2k$

$$n - \lceil \dfrac{n}{2} \rceil  = 2k - (k) = \lfloor \dfrac{2k}{2} \rfloor  = k $$

Both the LHS and RHS are $k$ therefore, the equality stands true.

Thus, we have shown  $n - \lceil \dfrac{n}{2} \rceil  = \lfloor \dfrac{n}{2} \rfloor$